Question

1/(1+x^2)从1到正无穷上求和是多少?

Answer 1

∑(n从1到正无穷) n(n+2)x^n =x∑(n从1到正无穷) n(n+2)x^(n-1) =x∑(n从1到正无穷)[(n+2)x^n]′ =x[∑(n从1到正无穷)(n+2)x^n]′ ∑(n从1到正无穷)(n+2)x^n =1/x[∑(n从1到正无穷)(n+2)x^(n+1)] =1/x∑(n从1到正无穷)[x^(n+2)]′ =1/x[∑(n从1到正无穷)x^(n+2)]′ =1/x[x3/(1-x)]′ =x(3-2x)/(1-x)2 原式=x[x(3-2x)/(1-x)2]′ =x(3-x)/(1-x)3

Answer 2

∫[1：+∞]dx/(1+x²)
=arctanx|[1：+∞]
=π/2 -π/4
=π/4